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∫ X − 3 X 2 + 2 X − 4 D X - Mathematics

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प्रश्न

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `

बेरीज

उत्तर

\[\int\left( \frac{x - 3}{x^2 + 2x - 4} \right)dx\]
\[x - 3 = A\frac{d}{dx}\left( x^2 + 2x - 4 \right) + B\]
\[x - 3 = A \left( 2x + 2 \right) + B\]
\[x - 3 = \left( 2 A \right) x + 2A + B\]

Comparing Coefficients of like powers of x

\[2A = 1\]
\[A = \frac{1}{2}\]
\[2A + B = - 3\]
\[2 \times \frac{1}{2} + B = - 3\]
\[B = - 4\]

\[Now, \int\left( \frac{x - 3}{x^2 + 2x - 4} \right)dx\]
\[ = \int\left( \frac{\frac{1}{2}\left( 2x + 2 \right) - 4}{x^2 + 2x - 4} \right)dx\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 2 \right) dx}{\left( x^2 + 2x - 4 \right)} - 4\int\frac{dx}{x^2 + 2x + 1 - 1 - 4}\]
\[ = \frac{1}{2}\int\frac{\left( 2x + 2 \right) dx}{\left( x^2 + 2x - 4 \right)} - 4\int\frac{dx}{\left( x + 1 \right)^2 - \left( \sqrt{5} \right)^2}\]
\[ = \frac{1}{2} \text{ log }\left| x^2 + 2x - 4 \right| - \frac{4}{2\sqrt{5}} \text{ log }\left| \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} \right| + C\]
\[ = \frac{1}{2} \text{ log }\left| x^2 + 2x - 4 \right| - \frac{2}{\sqrt{5}} \text{ log } \left| \frac{x + 1 - \sqrt{5}}{x + 1 + \sqrt{5}} \right| + C\]

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पाठ 19: Indefinite Integrals - Exercise 19.19 [पृष्ठ १०४]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.19 | Q 3 | पृष्ठ १०४

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